The radical vector of A-tilde 2 and its Euclidean slice

For $x=(x_1,x_2,x_3)$, direct expansion gives $B(x,x)=\frac{1}{2}((x_1-x_2{)}^2+(x_2-x_3{)}^2+(x_3-x_1{)}^2)$. It is nonnegative and vanishes exactly on $\mathbb{R}(1,1,1)$; matrix multiplication also gives $B(1,1,1)=0$, so this is precisely the radical. Each two-coordinate principal form is $a^2-ab+b^2=(a-b\mathrm{/}2{)}^2+3b^2\mathrm{/}4$, positive definite; the one-coordinate forms are $(1)$ and the empty case is vacuous. Their determinants are $3\mathrm{/}4$, and the full determinant is zero.

The slice relation is ${\phi }_1+{\phi }_2+{\phi }_3=1$, and [F3] gives its coordinate vertices $v_i$ and triangular closure. Put $H:=\{z\in {\mathbb{R}}^3:{\sum }_i{z}_i=0\}$. Every class in $U=V\mathrm{/}\mathbb{R}(1,1,1)$ has a unique representative in $H$, obtained by subtracting the mean of its coordinates, so $H\to U$ is a linear isomorphism. The direction space $K_{\delta }$ consists of functionals with coordinate tuple $y=(\psi (e_{s_1}),\psi (e_{s_2}),\psi (e_{s_3}))\in H$; under the representative identification, $\psi (\bar{z})={\sum }_i{y}_i{z}_i$, so $y$ represents $\psi$ using the standard dot product on $H$. For $z\in H$, the matrix in [F1] gives $Bz=(3\mathrm{/}2)z$; hence $b^{\mathrm{♭}}(z)$ is represented by $(3\mathrm{/}2)z$, and $b^{\mathrm{♭}}{}^{-1}\psi$ is represented by $z=(2\mathrm{/}3)y$. Therefore $b^{\ast }(\psi ,\psi )=B(z,z)=(2\mathrm{/}3){\sum }_i{y}_i^2$. Each difference $v_i-v_j$ has tuple with one $1$, one $-1$ and one zero, so its squared length is $4\mathrm{/}3$. All three sides therefore have length $2\mathrm{/}\sqrt{3}$ in the prescribed Euclidean slice metric.

The inward unit normals are $b^{\mathrm{♭}}({\bar{e}}_i)$; their pairings are $-1\mathrm{/}2$ for distinct indices by [F3]. The walls of the triangle thus meet at interior angle $\pi \mathrm{/}3$, and composing two line reflections gives rotation by $2\pi \mathrm{/}3$, of exact order three. The facet reflections generate the faithful action by [F3]. The relation among the normals is ${\sum }_i{b}^{\mathrm{♭}}({\bar{e}}_i)=b^{\mathrm{♭}}(\bar{\delta })=0$, with all coefficients one. Its positivity gives the bounded coordinate simplex directly: ${\phi }_i\ge 0$ and ${\sum }_i{\phi }_i=1$ bound each coordinate.

In the coordinate $A_2$ plane $x_1+x_2+x_3=0$, the closed root alcove is $x_1\ge x_2\ge x_3$ and $x_1-x_3\le 1$. Its vertices are the intersections of pairs of its three wall equations. The equations $x_1=x_2$ and $x_2=x_3$, together with the coordinate sum, give $(0,0,0)$. For $x_1=x_2$ and $x_1-x_3=1$, write $x_1=x_2=a$, $x_3=a-1$; then $3a-1=0$, giving $(1,1,-2)\mathrm{/}3$. For $x_2=x_3$ and $x_1-x_3=1$, write $x_2=x_3=b$, $x_1=b+1$; then $3b+1=0$, giving $(2,-1,-1)\mathrm{/}3$. Each point satisfies the remaining inequality. Each difference of two distinct vertices is, up to coordinate permutation and sign, $(2,-1,-1)\mathrm{/}3$, so its squared length is $(4+1+1)\mathrm{/}9=2\mathrm{/}3$. By [F4,F5] the root alcove and slice triangle are similar facet to facet, and the similarity from root alcove to slice has scale $\sqrt{2}$, since $(4\mathrm{/}3)\mathrm{/}(2\mathrm{/}3)=2$. It conjugates the corresponding reflection groups. This proves all the stated radical, wall, metric and comparison data without Choice.